Prove:
$$\lim_{\varepsilon \rightarrow 0^+} \int_0^\infty \frac{\varepsilon}{\varepsilon +x} \sin(\frac{1}{x}) = 0$$
I'm reviewing a past exam question. Could anyone point me in the right direction for starting this problem? I've tried simplifying the integral by integration by parts or u-substitution, but that hasn't worked for me.
Best Answer
Using $y=\frac{1}{x}$, $$\lim_{\varepsilon\to0^+}\int_0^\infty\frac{\sin\frac{1}{x}}{\varepsilon+x}dx=\lim_{\varepsilon\to0^+}\int_0^\infty\frac{1}{1+\varepsilon y}\frac{\sin y}{y}dy=\int_0^\infty\frac{\sin y}{y}dy=\frac{\pi}{2},$$where the penultimate $=$ uses the dominated convergence theorem. Thus$$\lim_{\varepsilon\to0^+}\int_0^\infty\frac{\varepsilon\sin\frac{1}{x}}{\varepsilon+x}dx=0.$$